Taylor Series
1. Find the degree 6 Taylor polynomial approximation for f (x) = sin x centered at 0.
Solution. We are looking for a polynomial P6 (x) = a0 + a1 x + a2 x2 + a3 x3 + a4 x4 + a5 x5 + a6 x6 such
(k)
(k)
that the k-th derivative P6 (0) is equal to the k-th derivative f (k) (0). The k-th derivative P6 (0) is
just equal to k!ak , so we want ak =
f (x)
f ′ (x)
f ′′ (x)
f ′′′ (x)
f (4) (x)
f (5) (x)
f (6) (x)
f (k) (0)
k! .
=
=
=
=
=
=
=
The derivatives of f (x) = sin x are:
sin x
cos x
− sin x
− cos x
sin x
cos x
− sin x
⇒
f (0)
⇒
f ′ (0)
⇒
f ′′ (0)
⇒ f ′′′ (0)
⇒ f (4) (0)
⇒ f (5) (0)
⇒ f (6) (0)
=
=
=
=
=
=
=
1
Therefore, a0 = 0, a1 = 1, a2 = 0, a3 = − 3!
, a4 = 0, a5 =
1
1
x − x3 + x5 .
3!
5!
0
1
0
−1
0
1
0
1
5! ,
and a6 = 0. So, P6 (x) =
2. (a) If you want to find a Taylor polynomial approximation a0 + a1 x + a2 x2 + · · · + an xn (centered at
0) to f (x), write a formula for the coefficient ak .
Solution. ak =
f (k) (0)
k!
. (You might wonder what happens when k = 0: 0! is defined to be 1, so
the formula still works.)
(b) If you want to find a Taylor polynomial approximation a0 + a1 (x− 3)+ a2 (x− 3)2 + · · ·+ an (x− 3)n
(centered at 3) to f (x), write a formula for the coefficient ak .
Solution. ak =
f (k) (3)
k! .
3. How do you think you would represent sin x as an infinite polynomial centered at 0? This is called the
Taylor series (rather than Taylor polynomial) generated by sin x about 0.
Solution. Based on the pattern we started to see in #1, it seems like we should get
x−
1 3
1
1
1
x + x5 − x7 + x9 − · · · .
3!
5!
7!
9!
4. What is the Taylor series generated by cos x about 0?
Solution. Using f (x) = cos x, we have
f (x)
f ′ (x)
f ′′ (x)
f ′′′ (x)
=
=
=
=
cos x
− sin x
− cos x
sin x
1
⇒
f (0)
⇒ f ′ (0)
⇒ f ′′ (0)
⇒ f ′′′ (0)
=
=
=
=
1
0
−1
0
After this, the derivatives repeat, and we continue to get 1, 0, −1, 0, 1, 0, −1, 0, . . .. So, our coefficients
1
, a5 = 0, and so on. Thus, the Taylor series should be
are a0 = 1, a1 = 0, a2 = − 21 !, a3 = 0, a4 = 4!
1−
1 2
1
1
1
x + x4 − x6 + x8 − · · · .
2!
4!
6!
8!
5. What is the Taylor series generated by ex about 0?
Solution. If f (x) = ex , then the k-th derivative f (k) (x) is always ex , so f (k) (0) = 1. Therefore, the
k-th coefficient ak in the Taylor series is
1
k! .
Thus, the Taylor series is 1 + x +
If we wanted to write this in summation notation, we would write
∞
X
xk
k=0
k!
x3
x4
x2
+
+
+ ··· .
2!
3!
4!
.
6. We hope that, by using “polynomials of infinite degree,” we end up with something that is not just
an approximation for our function but is actually equal to the function. We don’t really know if this
is true yet. Taking on faith that ex is actually equal to its Taylor expansion about 0, can you write a
power series expansion (or “infinite polynomial representation”) of:
2
(a) e−x ?
Solution. Since ex = 1 + x +
x2
2!
+
x3
3!
+
x4
4!
2
+ · · · , we can get e−x just by replacing all of the
2
x’s in the series for ex with −x2 : e−x = 1 − x2 +
2
e−x =
∞
X
(−x2 )k
k=0
this as
(b)
Z
∞
X
k!
(−1)k
k=0
x4
x6
x8
−
+
− · · · . In summation notation,
2!
3!
4!
. We often simplify (−x2 )k as [(−1)(x2 )]k = (−1)k x2k , so you might also see
x2k
.
k!
2
e−x dx?
2
Solution. If we believe that e−x = 1 − x2 +
x4
2!
x6
3!
+
x8
4!
− · · · , then it seems plausible that we
Z
2
1 x5
1
−
can integrate this using the reverse of the Power Rule to get e−x dx = C + x − x3 + ·
3
5 2!
7
9
1 x
1 x
·
+ ·
− · · · , where C is any constant.
7 3!
9 4!
−
7. (a) Write a general formula for the Taylor series of f (x) centered at 0.
(k)
Solution. We know it should look like a0 + a1 x + a2 x2 + a3 x3 + · · · where ak = f k!(0) . So, it is
∞
X
f ′′ (0) 2 f ′′′ (0) 3
f (k) (0) k
x +
x + · · · . In summation notation, it is
x .
f (0) + f ′ (0)x +
2!
3!
k!
k=0
(b) What if you wanted to center at 5?
Solution. Then, we would get a0 + a1 (x − 5) + a2 (x − 5)2 + a3 (x − 5)3 + · · · where ak =
In other words, we would get f (5) + f ′ (5)(x − 5) +
2
f (k) (5)
k! .
f ′′′ (5)
f ′′ (5)
(x − 5)2 +
(x − 5)3 + · · · .
2!
3!